Square roots of Jordan blocks

Question

It is known that for the Jordan block $$ J_p = \left(\overbrace{\begin{array}{cccc} \lambda & 1& \ldots & 0\\ 0 & \lambda& & 0\\ 0 & 0 & \ddots & 1 \\ 0 &0 &\ldots & \lambda \\ \end{array}}^{p}\right)$$ the following formulas are valid: $$ e^{J_p}= e^{\lambda} \sum_{r=0}^{p-1} \frac1{r!} I_r^{(p)}; \qquad {\mathrm{Ln}} {J_p}= I\; {\mathrm{Ln}} \lambda+ \sum_{r=1}^{p-1} \frac{(-1)^{r-1}}{r \lambda^r} I_r^{(p)}, $$ where $$ I_1^{(p)} = \left(\overbrace{\begin{array}{cccc} 0 & 1& \ldots & 0\\ 0 & 0& & 0\\ 0 & 0 & \ddots & 1 \\ 0 &0 &\ldots & 0 \\ \end{array}}^{p}\right),\quad I_2^{(p)} = \left(\overbrace{\begin{array}{cccccc} 0 & 0& 1& 0& \ldots & 0\\ 0 & 0& 0& 1&\ldots & 0\\ 0 & 0& 0& 0& \ddots & 1 \\ 0 & 0& 0& 0&\ldots & 0 \\ 0 & 0& 0& 0&\ldots & 0 \\ \end{array}}^{p}\right) \mbox{etc.} $$ My question is whether there is a similar formula for the square root. After some numerical experiments, I think yes. In particular, $$ \left(\begin{array}{cc} 1 & \frac{1}{2}\\ 0 & 1 \end{array}\right)^2= \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right), \quad \left(\begin{array}{ccc} 2 & \frac{1}{4} & - \frac{1}{64}\\ 0 & 2 & \frac{1}{4}\\ 0 & 0 & 2 \end{array}\right)^2 = \left(\begin{array}{ccc} 4 & 1 & 0\\ 0 & 4 & 1\\ 0 & 0 & 4 \end{array}\right), \quad \left(\begin{array}{ccc} 3 & \frac{1}{6} & - \frac{1}{216}\\ 0 & 3 & \frac{1}{6}\\ 0 & 0 & 3 \end{array}\right)^2= \left(\begin{array}{ccc} 9 & 1 & 0\\ 0 & 9 & 1\\ 0 & 0 & 9 \end{array}\right) $$ assumes the existence of a similar formula. Maybe someone knows it?

Answer 1

[EDITED] We have $J_p = \lambda I + I_1^{(p)}$. The binomial series $$ \sqrt{\lambda + x} = \sqrt{\lambda} \sum_{k=0}^\infty {1/2 \choose k} \left(\frac{x}{\lambda}\right)^k $$ applies. Since $(I_1^{(p)})^k = 0$ for $k > p$, it becomes a finite series $$ \sqrt{J_p} = \sum_{k=0}^p {1/2 \choose k} \lambda^{1/2-k} I_k^{(p)} $$ Here you can also write $$ {1/2 \choose k} = (-1)^{k+1} 2^{1-2k} \frac{(2k-2)! }{k! (k-1)!} \ \text{for}\ k \ge 1 $$